Q 12-14-207NEETNEET 2022Top questionMedium
The truth table for the given logic circuit is
Answer: (D) $\begin{array}{cc|c} A & B & C \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$
The upper NAND gets $A$ and $B$: $Y_1 = \overline{AB}$.
The lower NAND gets $\bar{A}$ (through the NOT gate) and $B$: $Y_2 = \overline{\bar{A}B}$.
The AND gate gives
$$C = \overline{AB}\cdot\overline{\bar{A}B} = (\bar{A} + \bar{B})(A + \bar{B}) = \bar{A}A + \bar{A}\bar{B} + A\bar{B} + \bar{B} = \bar{B}$$
So $C = 1, 0, 1, 0$ for $AB = 00, 01, 10, 11$, which is table (4).
Solution by Sreeraj P, M.Sc Physics